Solving integral (by substitution?)

Question

How do I solve the integral $\int \frac{1}{\sqrt{b-x^2}}$ where b is a constant ?

I know that $\int \frac{1}{\sqrt{1-x^2}} = \arcsin(x)$ , so I guess I have to substitute somehow clever. Can you give me a hint?

Answer 1

Use substitution $t=\sqrt{|b|}x$, then:

$$\int\frac{1}{\sqrt{b-x^2}}dx=\int \frac{\sqrt{|b|}}{\sqrt{b-bt^2}}dt=\int \frac{1}{\sqrt{1-t^2}}dt$$